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From the last hint, we have $latex BA_1=BO=R=CO=CA_1$ where $latex R$ is the circumradius of $latex ABC$. Hence,  $latex BOCA_1$ is a rhombus and $latex A_1C||BO$. Similarly, $latex BO||AC_1$ hence $latex A_1C||AC_1$. As $latex CAC_1A_1$ is a parallelogram, we also have $latex A_1C_1=AC$. We can similarly prove that $latex A_1B_1=AB$ and $latex B_1C_1=BC$. Thus $latex ABC\cong A_1B_1C_1$. This implies that it suffices to show that the centres of the two nine-point circles coincide. Remember that the nine-point centre is the midpoint of the line joining the circumcentre and the orthocentre.Claim $latex H$ is the circumcentre of $latex A_1B_1C_1$. Proof Note that $latex A_1H=R$ (as $latex A_1$ is the centre of $latex \odot BHC$ and $latex \odot BHC$ is a reflection of $latex \odot ABC$. Similarly, $latex B_1H=C_1H=R$.  Claim $latex O$ is the orthocentre of $latex A_1B_1C_1$. Proof As $latex A_1$ is the reflection of $latex O$ on $latex BC$, $latex A_1O\perp BC$. As $latex BC\parallel B_1C_1$, $latex A_1O\perp B_1C_1$. Similarly, $latex B_1O\perp C_1A_1$ and $latex C_1O\perp A_1B_1$. Thus $latex O$ is the orthocentre of $latex A_1B_1C_1$. 

Thus the centres of the nine-point circles of $latex ABC$ and $latex A_1B_1C_1$ coincide.

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